dsafew1231 2015-01-19 12:10
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tinyMCE Jquery和PHP表单提交

Am trying to submit stuff typed in tinyMCE using a Jquery function but am not certain how to do that and how the server side form-processor should accept the data.

This is what I have so far. Anyone see something that am missing here? I know am close. am just missing something.

Thanks in advance.

==THE HTML==

<form  onsubmit="save_tinyMCE_Stuff();" method="post" action="form_processor.php?action=save" >
  <textarea  class="classname" id="tme0" name="elm1" rows="25" cols="80" style="width: 100%" >Product Details Will appear Here. This information will be compiled by purchasing Department<br/>
  Some Text here
  </textarea><br /><input type="submit" name="save" value="Save" /><input type="reset" name="reset" value="Undo All" />
</form>

==THE SCRIPT IN HTML FILE==

save_tinyMCE_Stuff(){
  $dataString = tinymce.get('tme0').getContent();
  $zeurl = 'form_processor.php?action=save&data=' + $dataString;
  $.ajax({type: "POST",url: $zeurl,data: $dataString,cache: false,success: function(result){alert(result);}});
}

==THE PHP==

  form_processor.php
  <?php
if($_GET['action'] == 'save'){
  echo $_POST['WHAT NAME TO USE HERE'];
}
  ?>
  • 写回答

2条回答 默认 最新

  • doupao6011 2015-01-19 12:29
    关注

    You don't need to tinymce. Just submit textarea value.

    like this:

    function save_tinyMCE_Stuff() {
    
        $dataString = $('textarea').val();
        $zeurl = 'form_processor.php?action=save&data=' + $dataString;
        $.ajax({
            type: "POST",
            url: $zeurl,
            data: $dataString,
            cache: false,
            success: function(result) {
                alert(result);
            }
        });
    }
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论
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